Under the action of a given coulombic force, the acceleration of an electron is 2.5×10²²m/sec². Then, the magnitude of the acceleration of a proton under the action of same force is nearly

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

$1.6 \times 10^{- 19} m / \sec^{2}$

$9.1 \times 10^{31} m / \sec^{2}$

$1.5 \times 10^{19} m / \sec^{2}$

$1.6 \times 10^{27} m / \sec^{2}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{aligned} \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}} \Rightarrow \frac{2.5 \times 10^{22}}{a_{2}} = 1836 \\ \Rightarrow a_{2} = 1.5 \times 10^{19} m / \sec^{2} \end{aligned}$

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!